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Find the integral from negative five to negative four of π with respect to π₯ plus the integral from eight to negative five of π with respect to π₯, given that π is a constant.
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In this question, we are given two definite integrals and are asked to evaluate their sum.
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The integrand in both integrals is a constant denoted by π.
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The lower and upper limits in the first integral are negative five and negative four, respectively.
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And the lower and upper limits in the second integral are eight and negative five, respectively.
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Note that, in the first integral, the lower limit is less than the upper limit.
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And in the second integral, the lower limit is greater than the upper limit.
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In order to answer this question, we will make use of the following property.
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For any constant π, the integral from π to π of π with respect to π₯ is equal to π times π minus π.
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Recall that this property holds whether π is strictly less than π, π is equal to π, or π is strictly greater than π.
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Letting π equal negative five, π equal negative four, and π equal π in this property, we find that the integral from negative five to negative four of π with respect to π₯ is equal to π multiplied by negative four minus negative five, which simplifies to π multiplied by negative four plus five.
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This is equal to π.
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Since the property holds for π is greater than π, letting π equal eight, π equal negative five, and π equal π, we find that the integral from eight to negative five of π with respect to π₯ is equal to π multiplied by negative five minus eight, which simplifies to negative 13π.
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Therefore, the integral from negative five to negative four of π with respect to π₯ plus the integral from eight to negative five of π with respect to π₯ is equal to π minus 13π, which simplifies to negative 12π.
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This is our final answer.
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However, note that the method we have used to get to the answer is not the only method we couldβve used.
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Letβs quickly go through another method we could use to obtain the answer to the question.
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Recall that if we have two integrals who have the same integrand, with the property that the upper limit of the first integral is equal to the lower limit of the second integral, then their sum is equal to the integral of the same integrand from the lower limit of the first integral to the upper limit of the second integral.
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This property holds whether π is strictly less than π, π is equal to π, or π is strictly greater than π.
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We can use this property to answer the question.
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We can swap the order of the integrals in the sum we are asked to evaluate and rewrite the sum to be evaluated as the integral from eight to negative five of π with respect to π₯ plus the integral from negative five to negative four of π with respect to π₯.
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Now, we can see that the upper limit of the first integral is equal to the lower limit of the second integral.
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Therefore, using the property we have just described, we can rewrite the sum to be evaluated as the integral from eight to negative four of π with respect to π₯.
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Letting π equal eight, π equal negative four, and π equal π in the first property that we described, we evaluate this integral to be π multiplied by negative four minus eight, which simplifies to negative 12π, which is the same as the answer that we found previously.
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So we have seen two methods that can be used to find the sum of the definite integrals given to us in the question.
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And they both give us an answer of negative 12π.